Problem: $f(-1)=4\,$, $~f\,^\prime(-1)=2\,$, $~f\,^{\prime\prime}(-1)=-1\,$, and $~f\,^{\prime\prime\prime}(-1)=9\,$. What are the first four nonzero terms of the Taylor series, centered at $x=-1$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4+2(x+1)-\frac{1}{2}{{(x+1)}^{2}}+\frac{3}{2}{{(x+1)}^{3}}$ (Choice B) B $4+2(x-1)-\frac{1}{2}{{(x-1)}^{2}}+\frac{3}{2}{{(x-1)}^{3}}$ (Choice C) C $4+2(x+1)-\frac{1}{2}{{(x+1)}^{2}}+3{{(x+1)}^{3}}$ (Choice D) D $4+2(x-1)-\frac{1}{2}{{(x-1)}^{2}}+3{{(x-1)}^{3}}$ (Choice E) E $4+2(x+1)-{{(x+1)}^{2}}+9{{(x+1)}^{3}}$
Solution: We know the formula for the Taylor series centered at $~x=-1~$ for the function $~f\,$. $ f(-1)+f\,^\prime(-1)(x+1)+\frac{f\,^{\prime\prime}(-1)}{2!}{{(x+1)}^{2}}++...+\frac{{{f}^{(n)}}(-1)}{n!}{{(x+1)}^{n}}+...$ If we substitute the given function and derivative values, we obtain the following third-degree polynomial. $ T_3(x)=4+2(x+1)-\frac{1}{2!}{{(x+1)}^{2}}+\frac{9}{3!}{{(x+1)}^{3}}$ This simplifies to $ T_3(x)=4+2(x+1)-\frac{1}{2}{{(x+1)}^{2}}+\frac{3}{2}{{(x+1)}^{3}}\,$.